The Dehydration of Cyclohexanol to Form Cyclohexene
In order to dehydrate Cyclohexanol it is required that a dehydration
agent, in this case phosphoric acid, be added and for the mixture to
be distilled with the Cyclohexene being taken off between 343K and
363K. This reaction gives the formula:
In order to keep the experiment at a manageable size 0.1 mole of
Cyclohexanol was used along with 4cm3 of concentrated Phosphoric acid.
0.1 mole of Cyclohexanol x 100.2 = 10.2
100.2
0.962 = 10.4 cm3
However because the measuring cylinder only measured to the nearest
1cm3 ,10cm3 had to be used instead. This volume was placed into a 50cm3
pear shaped and the 4cm3 of phosphoric acid was added slowly while
shaking the flask. The flask was then attached to the rest of the
apparatus as shown below;
Because the boiling point of Cyclohexene is 356K and for water it
is373K it was decide best to reflux the mixture between the 343K and
368K or 70oC and 90oC as read off the thermometer. Because of this all
products collected in the vented receiver below 340K were discarded.
In order that water was not present inside the vented receiver or any
other component of the apparatus before the experiment began, all
components were cleaned with acitone. Once the experiment had been set
up the mixture was heated with a Bunsen burner to a temperature of
around 363K where it was kept in order to increase the speed at which
the Cyclohexene was collected in the vented receiver. The temperature
was measured with a thermometer with the mercury tip placed at the
entrance to the condenser in order to be most accurate for the
temperature of the gas be...
... middle of paper ...
...ol were used and therefore the yield should really be taken
on 0.096 moles instead, giving a yield of 72.5%
The measuring cylinder used to measure out the 0.096 moles of
Cyclohexanol was only accurate to the nearest 1cm3 therefore it is
quite possible that 9.5cm3 of Cyclohexanol was measured out which is
0.091 moles, giving a yield of 76% or based on 10.5 cm3, a yield of
69.0%. The balance was only accurate to the nearest 0.01g and
therefore the weight of Cyclohexene produced could instead have been
5.655g which, working on 0.091 moles of reactant, gives a yield of
76.7%. On the other hand the weight might have been 5.645 which,
working on 1.008 moles of reactant, gives a yield of 68.9%.
From this we can se that the maximum yield for the experiment is
around 77% which is reasonable for an experiment of this nature.
The experiment was not a success, there was percent yield of 1,423%. With a percent yield that is relatively high at 1,423% did not conclude a successful experiment, because impurities added to the mass of the actual product. There were many errors in this lab due to the product being transferred on numerous occasions as well, as spillage and splattering of the solution. Overall, learning how to take one product and chemically create something else as well as how working with others effectively turned out to be a
2-ethyl-1,3-hexanediol. The molecular weight of this compound is 146.2g/mol. It is converted into 2-ethyl-1-hydroxyhexan-3-one. This compounds molecular weight is 144.2g/mol. This gives a theoretical yield of .63 grams. My actual yield was .42 grams. Therefore, my percent yield was 67%. This was one of my highest yields yet. I felt that this was a good yield because part of this experiment is an equilibrium reaction. Hypochlorite must be used in excess to push the reaction to the right. Also, there were better ways to do this experiment where higher yields could have been produced. For example PCC could have been used. However, because of its toxic properties, its use is restricted. The purpose of this experiment was to determine which of the 3 compounds was formed from the starting material. The third compound was the oxidation of both alcohols. This could not have been my product because of the results of my IR. I had a broad large absorption is the range of 3200 to 3500 wavenumbers. This indicates the presence of an alcohol. If my compound had been fully oxidized then there would be no such alcohol present. Also, because of my IR, I know that my compound was one of the other 2 compounds because of the strong sharp absorption at 1705 wavenumbers. This indicates the presence of a carbonyl. Also, my 2,4-DNP test was positive. Therefore I had to prove which of the two compounds my final product was. The first was the oxidation of the primary alcohol, forming an aldehyde and a secondary alcohol. This could not have been my product because the Tollen’s test. My test was negative indicating no such aldehyde. Also, the textbook states that aldehydes show 2 characteristic absorption’s in the range of 2720-2820 wavenumbers. No such absorption’s were present in my sample. Therefore my final product was the oxidation of the secondary alcohol. My final product had a primary alcohol and a secondary ketone
In this experiment there were eight different equations used and they were, molecular equation, total ionic equation, net ionic equation, calculating the number of moles, calculating the theoretical yield and limiting reagent, calculating the mass of〖PbCrO〗_4, calculating actual yield, calculating percent yield (Lab Guide pg.83-85).
Wittig reactions allow the generation of an alkene from the reaction between an aldehyde/ketone and a ylide (derived from phosphonium salt).The mechanism for the synthesis of trans-9-(2-phenylethenyl) anthracene first requires the formation of the phosphonium salt by the addition of triphenylphosphine and alkyl halide. The phosphonium halide is produced through the nucleophilic substitution of 1° and 2° alkyl halides and triphenylphosphine (the nucleophile and weak base) 4 An example is benzyltriphenylphosphonium chloride which was used in this experiment. The second step in the formation of the of the Wittig reagent which is primarily called a ylide and derived from a phosphonium halide. In the formation of the ylide, the phosphonium ion in benzyltriphenylphosphonium chloride is deprotonated by the base, sodium hydroxide to produce the ylide as shown in equation 1. The positive charge on the phosphorus atom is a strong EWG (electron-withdrawing group), which will trigger the adjacent carbon as a weak acid 5 Very strong bases are required for deprotonation such as an alkyl lithium however in this experiment 50% sodium hydroxide was used as reiterated. Lastly, the reaction between ylide and aldehyde/ketone produces an alkene.3
The theoretical mass of ethyl salicylate to be obtained was 1.667 grams and the actual mass obtained was 0.600 g.
The alcohol starting material, 2-methylcyclohexanol, was dehydrated through an E1 elimination by using of phosphoric acid as a catalyst. After a purification by simple distillation, which removed the alkene product and the by-product water from the reaction mixture, the methylcyclohexene products were analyzed by percent yield, boiling point, IR spectroscopy, and two chemical tests, Br2 in CCl4 and Jones test. By performing the simple distillation using pyrolysis, 85% of phosphoric acid and 2-methylcyclohexanol were added into the boiling flask, where the product from the collecting flask was condensed by the ice, and washed with the saturated sodium chloride. The weight of the product was determined and the percent yield of the product was
10cm3 of 1 molar solution. I will use 3 of each solution to ensure that
steps to obtain pure xylose. Moreover, the yield of xylitol is only 50 % of
this is the best volume to use as it is about ¾ of a test tube full,
Results & Discussion:The actual, theoretical, and percent yield of sodium chloride (NaCl) was found to be 1.14g, .700g, and 61.4%,
Figure 1: Simple batch homogenous reactor. [Fogler, H. S. (2010, November 22). Essentials of Chemical Reaction Engineering: Mole Balances. Retrieved April 24, 2014, from Pearson Education: http://www.informit.com/articles/article.aspx?p=1652026&seqNum=3]
Since I am making a 0.1 Mconcentration, I will need 0.001 moles of each sugar.
Toluene hydrodealkylation or hydrodealkylation of toluene (HDA) is a process that used to produce benzene. The reaction occurs as:
remaining 20 percent is due to other gasses that are present in very small amounts? (Murck,